I'm trying to reverse a string using the code:
puts("Hi now it's going to be done!")
string = gets.chomp.to_s
i = string.length
while i >= 0
puts(string[i])
i = i - 1
end
It prints the string in backward order, but each word is on a single line. How can I keep all of them on a single line?
puts adds a newline to the end of the output if one isn't already present.
print does not. So do this:
while i >=0
print string[i]
i=i-1
end
puts
The final puts is because you want any further printing to be on a new line.
Try this:
"Hi now it's going to be done!".chars.inject([]) { |s, c| s.unshift(c) }.join
Or This is a little easier to follow:
string = 'Hi now it's going to be done!'
string.reverse!
Related
I'm attempting to write a function that takes a string and returns it with all vowels removed. Below is my code.
def vowel(str)
result = ""
new = str.split(" ")
i = 0
while i < new.length
if new[i] == "a"
i = i + 1
elsif new[i] != "a"
result = new[i] + result
end
i = i + 1
end
return result
end
When I run the code, it returns the exact string that I entered for (str). For example, if I enter "apple", it returns "apple".
This was my original code. It had the same result.
def vowel(str)
result = ""
new = str.split(" ")
i = 0
while i < new.length
if new[i] != "a"
result = new[i] + result
end
i = i + 1
end
return result
end
I need to know what I am doing wrong using this methodology. What am I doing wrong?
Finding the bug
Let's see what's wrong with your original code by executing your method's code in IRB:
$ irb
irb(main):001:0> str = "apple"
#=> "apple"
irb(main):002:0> new = str.split(" ")
#=> ["apple"]
Bingo! ["apple"] is not the expected result. What does the documentation for String#split say?
split(pattern=$;, [limit]) → anArray
Divides str into substrings based on a delimiter, returning an array of these substrings.
If pattern is a String, then its contents are used as the delimiter when splitting str. If pattern is a single space, str is split on whitespace, with leading whitespace and runs of contiguous whitespace characters ignored.
Our pattern is a single space, so split returns an array of words. This is definitely not what we want. To get the desired result, i.e. an array of characters, we could pass an empty string as the pattern:
irb(main):003:0> new = str.split("")
#=> ["a", "p", "p", "l", "e"]
"split on empty string" feels a bit hacky and indeed there's another method that does exactly what we want: String#chars
chars → an_array
Returns an array of characters in str. This is a shorthand for str.each_char.to_a.
Let's give it a try:
irb(main):004:0> new = str.chars
#=> ["a", "p", "p", "l", "e"]
Perfect, just as advertised.
Another bug
With the new method in place, your code still doesn't return the expected result (I'm going to omit the IRB prompt from now on):
vowel("apple") #=> "elpp"
This is because
result = new[i] + result
prepends the character to the result string. To append it, we have to write
result = result + new[i]
Or even better, use the append method String#<<:
result << new[i]
Let's try it:
def vowel(str)
result = ""
new = str.chars
i = 0
while i < new.length
if new[i] != "a"
result << new[i]
end
i = i + 1
end
return result
end
vowel("apple") #=> "pple"
That looks good, "a" has been removed ("e" is still there, because you only check for "a").
Now for some refactoring.
Removing the explicit loop counter
Instead of a while loop with an explicit loop counter, it's more idiomatic to use something like Integer#times:
new.length.times do |i|
# ...
end
or Range#each:
(0...new.length).each do |i|
# ...
end
or Array#each_index:
new.each_index do |i|
# ...
end
Let's apply the latter:
def vowel(str)
result = ""
new = str.chars
new.each_index do |i|
if new[i] != "a"
result << new[i]
end
end
return result
end
Much better. We don't have to worry about initializing the loop counter (i = 0) or incrementing it (i = i + 1) any more.
Avoiding character indices
Instead of iterating over the character indices via each_index:
new.each_index do |i|
if new[i] != "a"
result << new[i]
end
end
we can iterate over the characters themselves using Array#each:
new.each do |char|
if char != "a"
result << char
end
end
Removing the character array
We don't even have to create the new character array. Remember the documentation for chars?
This is a shorthand for str.each_char.to_a.
String#each_char passes each character to the given block:
def vowel(str)
result = ""
str.each_char do |char|
if char != "a"
result << char
end
end
return result
end
The return keyword is optional. We could just write result instead of return result, because a method's return value is the last expression that was evaluated.
Removing the explicit string
Ruby even allows you to pass an object into the loop using Enumerator#with_object, thus eliminating the explicit result string:
def vowel(str)
str.each_char.with_object("") do |char, result|
if char != "a"
result << char
end
end
end
with_object passes "" into the block as result and returns it (after the characters have been appended within the block). It is also the last expression in the method, i.e. its return value.
You could also use if as a modifier, i.e.:
result << char if char != "a"
Alternatives
There are many different ways to remove characters from a string.
Another approach is to filter out the vowel characters using Enumerable#reject (it returns a new array containing the remaining characters) and then join the characters (see Nathan's answer for a version to remove all vowels):
def vowel(str)
str.each_char.reject { |char| char == "a" }.join
end
For basic operations like string manipulation however, Ruby usually already provides a method. Check out the other answers for built-in alternatives:
str.delete('aeiouAEIOU') as shown in Gagan Gami's answer
str.tr('aeiouAEIOU', '') as shown in Cary Swoveland's answer
str.gsub(/[aeiou]/i, '') as shown in Avinash Raj's answer
Naming things
Cary Swoveland pointed out that vowel is not the best name for your method. Choose the names for your methods, variables and classes carefully. It's desirable to have a short and succinct method name, but it should also communicate its intent.
vowel(str) obviously has something to do with vowels, but it's not clear what it is. Does it return a vowel or all vowels from str? Does it check whether str is a vowel or contains a vowel?
remove_vowels or delete_vowels would probably be a better choice.
Same for variables: new is an array of characters. Why not call it characters (or chars if space is an issue)?
Bottom line: read the fine manual and get to know your tools. Most of the time, an IRB session is all you need to debug your code.
I should use regex.
str.gsub(/[aeiou]/i, "")
> string= "This Is my sAmple tExt to removE vowels"
#=> "This Is my sAmple tExt to removE vowels"
> string.delete 'aeiouAEIOU'
#=> "Ths s my smpl txt t rmv vwls"
You can create a method like this:
def remove_vowel(str)
result = str.delete 'aeiouAEIOU'
return result
end
remove_vowel("Hello World, This is my sample text")
# output : "Hll Wrld, Ths s my smpl txt"
Live Demo
Assuming you're trying to learn about the basics of programming, rather than finding the quickest one-liner to do this (which would be to use a regular expression as Avinash has said), you have a number of problems with your code you need to change.
new = str.split(" ")
This line is likely the culprit, because it splits the string based on spaces. So your input string would have to be "a p p l e" to have the effect you're looking for.
new = str.split("")
You should also remove the duplicate i = i+1 once you've changed that.
As others have already identified the problems with the OP's code, I will merely suggest an alternative; namely, you could use String#tr:
"Now is the time for all good people...".tr('aeiouAEIOU', '')
#=> "Nw s th tm fr ll gd ppl..."
If regex is not allowed, you can do it this way:
def remove_vowels(string)
string.split("").delete_if { |letter| %w[a e i o u].include? letter }.join
end
I was trying to replace a given pattern with some string in a object which is of class Nokogiri::XML::Text using gsub in ruby..see the below command
#str is of class Nokogiri::XML::Text
str.content = str.content.gsub(pattern,replacing_word)
Now I wanted to print something with each replacement and also wanted to know the number of replacement that gsub did so I wrote the below command
count = 0
str.content = str.content.gsub(pattern,replacing_word) { count += 1
puts "some text"}
The above command is replacing the given pattern with the replacing_string but the body part of gsub is not getting executed, Neither any print statement nor any increment operation on count is happening.
Even I tried with just puts statement in the body then also it is not printing, although this gsub is doing many replacements in "str"
Let me know if anyone knows what is the issue with this gsub command
try the following code snippet instead of yours,
content = content.gsub(pattern) {|m| count +=1; m.replace(replacing_word)}
My Testing Code
#!/usr/bin/env ruby
pattern = "a";
content = "abaccaa"
replacing_word = "z"
count = 0
content = content.gsub(pattern) {|m| count +=1; m.replace(replacing_word)}
puts content;
puts count;
Output
zbzcczz
4
--SJ
I'm currently going through some programs to learn Ruby. I've been playing around with a palindrome program for a bit, though no matter the input (a palindrome) I end up on else.
Here is some of the code I've been trying:
print "enter a string:\n"
string = gets
if string.reverse == string
print "it's a palindrome"
else
print "not a palindrome.\n"
end
Any help/advice is greatly appreciated.
The newline character is not being deleted from the string. Try this code:
print "enter a string:\n"
string = gets.chomp
if string.reverse == string
print "it's a palindrome"
else
print "not a palindrome.\n"
end
Here is some more explanation:
>> string = gets
racecar # input string
=> "racecar\n"
>> "racecar\n" == "racecar\n".reverse # "racecar\n" is not a palindrome with newline character
=> false
>> string = gets.chomp # chomp method deletes newline character
racecar
=> "racecar"
>> "racecar" == "racecar".reverse # "racecar" without a newline character is a palindrome
=> true
Learn how Ruby's puts works: It's like print, only smarter.
If a string ends with "\n", it prints it as is. If it doesn't end with "\n", it prints the line and adds "\n". Either way, you're guaranteed to have new-line added.
Knowing that, consider this:
puts "enter a string:"
string = gets
if string.reverse == string
puts "it's a palindrome"
else
puts "not a palindrome."
end
As a result, no new-lines need to be added to the strings. puts is the standard method for outputting lines to files and the console in Ruby.
The following statement will return true if it's a palindrome and false otherwise:
string == string.reverse
I need to make a method that repeats a given word but I think I am designing it wrong. I need spaces between the words, what am I missing here?
def repeat(word, repeats=2)
sentence = word.to_s * repeats
return sentence
end
Of course, you are missing spaces.
You could have done it like this:
def repeat(word, repeats = 2)
Array.new(repeats, word).join(" ")
end
You can write the code as below :
def repeat(word, repeats=2)
([word] * repeats).join(" ")
end
repeat("Hello",4)
# => "Hello Hello Hello Hello"
Here's one closer to your approach and without using a temporary array:
def repeat(word, repeats=2)
("#{word} " * repeats).chop
end
"#{word} " converts word to a string using string interpolation and appends a space
… * repeats creates a string containing repeats copies
.chop returns the string with the last character removed (the trailing space)
Not having to create an array makes the code a bit faster.
w = 'kokot'
n = 13
n.times.map { w.each_char.to_a.shuffle.join }.tap { |a, _| a.capitalize! }.join(' ') << ?.
I'm playing around with Ruby to do some file versioning for me. I have a string 2.0.0.65 . I split it up, increment the build number (65 --> 66) then I want to replace the 65 with the 66. In this replace though, I only want to replace the last match of the string. What's the best way in Ruby to do this?
version_text = IO.read('C:\\Properties')
puts version_text
version = version_text.match(/(\d+\.\d+\.\d+\.\d+)/)[1]
puts version
build_version = version.split('.')[3]
puts build_version
incremented_version = build_version.to_i + 1
puts incremented_version`
...
If you just want to increment the integer at the very end of a string then try this:
s = '2.0.0.65'
s.sub(/\d+\Z/) {|x| x.to_i + 1} # => '2.0.0.66'
You can do something like this:
parts = "2.0.0.65".split('.')
parts[3] = parts[3].to_i + 1
puts parts.join(".")
output:
2.0.0.66
This gives you more control over just using a string replacement method, as now you can increment other parts of the version string if needed more easily.
Once you have the string with the build number, you only need to use 'succ' method
'2.0.0.65'.succ()
Which gives you the string
'2.0.0.66'
sample = '2.0.0.65'
def incr_version(version)
parts = version.split('.')
parts[-1] = parts[-1].to_i + 1
parts.join('.')
end
incr_version(sample) # => '2.0.0.66'
For fun, if you want to increment the last integer in any string you could do this:
str = "I have 3 cats and 41 rabbits"
str.reverse.sub(/\d+/){ |s| (s.reverse.to_i+1).to_s.reverse }.reverse
#=> "I have 3 cats and 42 rabbits"
This is only valid when you modify your regex to match the reversed version of the text.
More generally, you can do this:
class String
# Replace the last occurrence of a regex in a string.
# As with `sub` you may specify matches in the replacement string,
# or pass a block instead of the replacement string.
# Unlike `sub` the captured sub-expressions will be passed as
# additional parameters to your block.
def rsub!(pattern,replacement=nil)
if n=rindex(pattern)
found=match(pattern,n)
self[n,found[0].length] = if replacement
replacement.gsub(/\\\d+/){ |s| found[s[1..-1].to_i] || s }
else
yield(*found).to_s
end
end
end
def rsub(pattern,replacement=nil,&block)
dup.tap{ |s| s.rsub!(pattern,replacement,&block) }
end
end
str = "I have 3 cats and 41 rabbits"
puts str.rsub(/(?<=\D)(\d+)/,'xx')
#=> I have 3 cats and xx rabbits
puts str.rsub(/(?<=\D)(\d+)/,'-\1-')
#=> I have 3 cats and -41- rabbits
puts str.rsub(/(?<=\D)(\d+)/){ |n| n.to_i+1 }
#=> I have 3 cats and 42 rabbits
Note that (as with rindex) because the regex search starts from the end of the string you may need to make a slightly more complex regex to force your match to be greedy.